Problem: The lifespans of lizards in a particular zoo are normally distributed. The average lizard lives $2.6$ years; the standard deviation is $0.6$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a lizard living longer than $2$ years.
Solution: $2.6$ $2$ $3.2$ $1.4$ $3.8$ $0.8$ $4.4$ $68\%$ $16\%$ $16\%$ We know the lifespans are normally distributed with an average lifespan of $2.6$ years. We know the standard deviation is $0.6$ years, so one standard deviation below the mean is $2$ years and one standard deviation above the mean is $3.2$ years. Two standard deviations below the mean is $1.4$ years and two standard deviations above the mean is $3.8$ years. Three standard deviations below the mean is $0.8$ years and three standard deviations above the mean is $4.4$ years. We are interested in the probability of a lizard living longer than $2$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $68\%$ of the lizards will have lifespans within 1 standard deviation of the average lifespan. The remaining $32\%$ of the lizards will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({16\%})$ will live less than $2$ years and the other half $({16\%})$ will live longer than $3.2$ years. The probability of a particular lizard living longer than $2$ years is ${68\%} + {16\%}$, or $84\%$.